Let $G$ be a topological group that acts (topologically) on a topological space $V=\mathbb Q_p^n$. Moreover, let us assume that this action is linear, in other words, the action gives a linear representation $\rho: G \to GL(V)$.
I wonder whether the openness of stabilizers $G_x$ for all $x \in X$ implies that $\rho(G)$ is discrete in $GL(V)$, and vice versa. I tried to prove but still, I am not sure that this is true.
The motivation for this question comes from $p$-adic Galois representations. For $p$-adic field $K$, it is known that the $\overline K$-admissibility of a $p$-adic representation $V$ of $G_K$ (the absolute Galois group of $K$) is equivalent to the discreteness of an action $G_K$ on $V$. And it is stated that this is implied once if one proves that $G_{K,x}$ is open for arbitrary $x \in V$.
The absolute Galois group is compact, so we will assume $G$ is compact in what follows.
Let $v_1, \ldots, v_n \in V$ be a basis; this induces an homeomorphism $\Psi$ from $\mathrm{GL}(V)$ to the space of frames $\mathrm{F}(V)$ given by $$T \overset{\Psi}\longmapsto (Tv_1, \ldots Tv_n).$$
Let $\operatorname{orb}_G(v)$ and $\operatorname{stab}_G(v)$ denote the orbit and stabilizer of a vector $v$, respectively. It suffices to prove that the orbit of each basis vector is discrete since $$ \rho(G) \overset{\Psi}\hookrightarrow \prod_{i=1}^{n} \operatorname{orb}_G(v_i). $$ Consider the map $\rho_i \colon G \to V$ given by $\rho_i(g) = gv_i$. Let $u \in \rho_i(G)$, meaning that $u = gv_i$ for some $g \in G$. Then the fiber over $u$ is given by $$\rho_i^{-1}(\{u\}) = g\operatorname{stab}_G(v_i).$$
Since the fibers of $\rho_i$ are precisely the cosets of the stabilizer, the universal property of the quotient yields a continuous bijection $\widetilde{\rho}_i \, \colon G / \operatorname{stab}_G(v_i) \to \operatorname{orb}_G(v_i)$, which is actually a homeomorphism, since $G$ is compact and $V$ is Hausdorff.
To wrap up, note that the quotient $G / \operatorname{stab}_G(v_i)$ is discrete, since $\operatorname{stab}_G(v_i)$ is open.